Explain why there are two complex numbers z such that $|z| = 1$ and that satisfy the equation $|z| = |z-1|.$

Question

I must find both such complex solutions and express them in Euler form and usual form.

So it's been a while since I've touched the imaginary/real plane. However, from what I remember, $z = a + bi$. If $|z| = 1$ we can assign specific values to a and b.. what would these values be?.. And how would we use the result to find specific $z$ such that $|z| = |z-1|$?

Answer 1

Hint: Think two-dimensionally. Our $z$ is equidistant from $(0,0)$ and $(1,0)$. Geometry shows that $z$ must lie on the perpendicular bisector of the line segment that joins $(0,0)$ to $(1,0)$. In complex number terms, the real part of $z$ is $\frac{1}{2}$.

Answer 2

Hints:

Write $\;z=x+iy\;,\;x,y\;\Bbb R\;$ , so that

$$\begin{cases}|z|=1\iff x^2+y^2=1\\{}\\|z|=|z-1|\iff x^2+y^2=(x-1)^2+y^2\end{cases}\;\;\;\implies$$

$$2x=1\iff x=\frac12\;\;$$

So now just substitute in the first equation above and...

Answer 3

One way to think of this geometrically is that the function $f(z) = |z - p|$ gives you the distance of a complex number $z$ from the point $p$. If we have the equality $$ |z - 0| = |z| = |z-1| $$ this tells you that we are looking for points $z$ that are equidistant from both the origin, 0, and the point $1 \in \mathbb{C}$.

It isn't too hard to see that this is the line of points $x + iy \in \mathbb{C}$ such that $x = \frac{1}{2}$ (draw a picture).

From this, it follows that you're looking for the intersection of this line with the circle $|z| = 1$, from which you should be able to determine your solution.

Answer 4

I think the mistake in your reasoning is here:

If |z|=1 we can assign specific values to a and b. what would these values be?

In fact, knowing that |z|=1 does not allow you to determine values of a and b. All that |z|=1 tells you is that z is exactly one unit from the origin -- in other words, it lies on a circle of radius 1, centered at 0.

Likewise, the condition that |z|=|z-1| means that z is equidistant from the two points 0 and 1, which means it lies on a line through 1/2 and perpendicular to the real axis.

Finishing touch: In how many points does that line cross the circle?

Answer 5

Note that $|z|=1$ if and only if $z=e^{it}$ for some real $t$. So we have to solve $1=|e^{it}-1|$. But that's easy: $$1=|e^{it}-1| \iff 1 =|e^{it}-1|^2$$ and $$|e^{it}-1|^2 =|(\cos t -1)+i\sin t|^2 =(\cos t - 1)^2 + \sin^2 t$$ $$= \cos^2 t -2\cos t + 1 + \sin^2 t = 2 - 2\cos t$$ This means we have $$1 = 2 - 2\cos t \iff \cos t = \frac12$$ which means that $$\sin t = \pm \frac{\sqrt 3}{2}$$ and so $$z = e^{it} = \cos t +i\sin t = \boxed{\frac12 \pm i \frac{\sqrt 3}{2}}$$